# Difference between revisions of "Specimen Cyprus Seniors Provincial/2nd grade/Problem 4"

(problem 4) |
(added solution) |
||

Line 8: | Line 8: | ||

== Solution == | == Solution == | ||

+ | Since <math>\rho_1, \rho_2 \neq -1</math>, we can multiply both sides by <math>x+1</math>, and <math>\rho_{1}, \rho_{2}</math> will still satisfy the equation: | ||

+ | <math>(x^2-x+1)(x+1)=(0)(x+1)</math> | ||

+ | <math>x^3 + 1 = 0</math> | ||

+ | |||

+ | <math>x^3 = -1</math> | ||

+ | |||

+ | Thus, <math>\rho_{1}^3=\rho_{2}^3 = -1</math> as desired <math>\boxed{\mathbb{Q.E.D.}}</math>. | ||

+ | |||

+ | |||

+ | The third roots of <math>-1 = cis(180^\circ)</math> are <math>cis(60^\circ)</math>, <math>cis(-60^\circ)</math>, and <math>-1</math>. | ||

+ | |||

+ | Since <math>\rho_1, \rho_2 \neq -1</math>, we have <math>\rho_{1} = cis(60^\circ)</math> and <math>\rho_{2} = cis(-60^\circ)</math>. (Switching the two values will not affect the result). | ||

+ | |||

+ | Note that <math>\rho_{1}^6=\rho_{2}^6 = 1</math>, and that <math>2006 = 6(334)+2</math> | ||

+ | |||

+ | So, <math>\rho_{1}^{2006} + \rho_{2}^{2006} = \rho_{1}^{2} + \rho_{2}^{2} = cis(120^\circ) + cis(-120^\circ) = </math> <math>\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i \right) + \left(-\frac{1}{2} - \frac{\sqrt{3}}{2}i \right) = \boxed{-1}</math> | ||

---- | ---- |

## Latest revision as of 23:28, 22 May 2009

## Problem

If are the roots of equation then:

a) Prove that and

b) Calculate the value of: .

## Solution

Since , we can multiply both sides by , and will still satisfy the equation:

Thus, as desired .

The third roots of are , , and .

Since , we have and . (Switching the two values will not affect the result).

Note that , and that

So,